Track / Stage 6 / 13 of 400 #1013 of 1964

Problem 1013

National olympiad, first round

Geometry Difficulty 6.0 Prove it

2. A closed broken line passing through each vertex exactly once is drawn along the edges of a convex polyhedron with 2003 vertices. Prove that in each of the parts into which this broken line divides the surface of the polyhedron, the number of faces with an odd number of sides is odd.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

2. Let's choose any of the resulting parts. Consider the sum $a_{1}+a_{2}+\ldots+a_{n}$ , where $a_{i}$ is the number of sides of the $i$ -th face.

Each edge of the polyhedron that the broken line does not pass through is counted twice in this sum, and therefore the parity of the sum does not depend on the number of such edges (fact 23).

Each edge that the broken line passes through is included in the sum exactly once. There are 2003 such edges, so the entire sum is odd.

If the number of faces with an odd number of sides were even, then the considered sum would also be even. Therefore, this number is odd.

Comments. $1^{\circ}$ . Provide an example of such a polyhedron and such a broken line.

$2^{\circ}$ . Compare this problem with the well-known problem about Martians. Martians have an arbitrary number of hands. One day, all Martians held hands so that all hands were occupied. Prove that the number of Martians with an odd number of hands is even.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.